ref: c172881606bfa194399a0b181b419a34488331d6
dir: /sys/src/ape/lib/ap/math/erf.c/
#include <math.h> #include <errno.h> /* C program for floating point error function erf(x) returns the error function of its argument erfc(x) returns 1 - erf(x) erf(x) is defined by ${2 over sqrt(pi)} int from 0 to x e sup {-t sup 2} dt$ the entry for erfc is provided because of the extreme loss of relative accuracy if erf(x) is called for large x and the result subtracted from 1. (e.g. for x= 10, 12 places are lost). There are no error returns. Calls exp. Coefficients for large x are #5667 from Hart & Cheney (18.72D). */ #define M 7 #define N 9 static double torp = 1.1283791670955125738961589031; static double p1[] = { 0.804373630960840172832162e5, 0.740407142710151470082064e4, 0.301782788536507577809226e4, 0.380140318123903008244444e2, 0.143383842191748205576712e2, -.288805137207594084924010e0, 0.007547728033418631287834e0, }; static double q1[] = { 0.804373630960840172826266e5, 0.342165257924628539769006e5, 0.637960017324428279487120e4, 0.658070155459240506326937e3, 0.380190713951939403753468e2, 0.100000000000000000000000e1, 0.0, }; static double p2[] = { 0.18263348842295112592168999e4, 0.28980293292167655611275846e4, 0.2320439590251635247384768711e4, 0.1143262070703886173606073338e4, 0.3685196154710010637133875746e3, 0.7708161730368428609781633646e2, 0.9675807882987265400604202961e1, 0.5641877825507397413087057563e0, 0.0, }; static double q2[] = { 0.18263348842295112595576438e4, 0.495882756472114071495438422e4, 0.60895424232724435504633068e4, 0.4429612803883682726711528526e4, 0.2094384367789539593790281779e4, 0.6617361207107653469211984771e3, 0.1371255960500622202878443578e3, 0.1714980943627607849376131193e2, 1.0, }; double erfc(double); double erf(double arg) { int sign; double argsq; double d, n; int i; errno = 0; sign = 1; if(arg < 0) { arg = -arg; sign = -1; } if(arg < 0.5) { argsq = arg*arg; for(n=0,d=0,i=M-1; i>=0; i--) { n = n*argsq + p1[i]; d = d*argsq + q1[i]; } return sign*torp*arg*n/d; } if(arg >= 10) return sign; return sign*(1 - erfc(arg)); } double erfc(double arg) { double n, d; int i; errno = 0; if(arg < 0) return 2 - erfc(-arg); if(arg < 0.5) return 1 - erf(arg); if(arg >= 10) return 0; for(n=0,d=0,i=N-1; i>=0; i--) { n = n*arg + p2[i]; d = d*arg + q2[i]; } return exp(-arg*arg)*n/d; }